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想要通过的同学就想办法让第二个样例输入“ 1 4 3 1”In Reply To:这题的测试数据……………… 首先第一个数字前应该有空格 其次SPECIAL JUDGE是必须的 可惜现在还没有 Posted by:rruucc at 2003-09-13 00:32:00 > //附上程序 简单的迭代实现 不过试了很多参数才AC > #include<stdio.h> > #include<string.h> > #define MaxN 21 > #define Profit 1.01 > > int n; > double map[MaxN][MaxN]; > double arb[MaxN][MaxN]; > int way[MaxN][MaxN][MaxN]; > int from,to,len; > > int init() > { int i,j; > if (scanf("%d",&n)==EOF) return 0; > for (i=0; i<n; i++) > for (j=0; j<n; j++) > {if (i==j) map[i][j]=1; else scanf("%lf",&map[i][j]); > arb[i][j]=map[i][j]; way[i][j][0]=j; > } > return 1; > } > > void search() > { int i,j,k,h; > double tmp[MaxN][MaxN]; > int tmpw[MaxN][MaxN][MaxN]; > from=-1; to=-1; len=-1; > for (j=0; j<n; j++) > for (k=0; k<n; k++) > if (arb[j][k]*map[k][j]>Profit) > {from=j; to=k; len=0; return;} > for (i=1; i<=n-2; i++) > {memcpy(tmp,arb,sizeof(arb)); > memcpy(tmpw,way,sizeof(way)); > for (j=0; j<n; j++) > for (k=0; k<n; k++) arb[j][k]=-1; > for (j=0; j<n; j++) > for (h=0; h<n; h++) > for (k=0; k<n; k++) > if (tmp[j][k]*map[k][h]>arb[j][h]) > {arb[j][h]=tmp[j][k]*map[k][h]; > memcpy(way[j][h],tmpw[j][k],sizeof(tmpw[j][k])); > way[j][h][i]=h; > } > for (j=0; j<n; j++) > for (k=0; k<n; k++) > if (arb[j][k]*map[k][j]>Profit) > {from=j; to=k; len=i; return;} > } > } > > void show() > { int k; > if (len==-1) printf("no arbitrage sequence exists\n"); > else > {printf(" %d",from+1); > for (k=0; k<=len; k++) printf(" %d",way[from][to][k]+1); > printf(" %d\n",from+1); > } > } > > int main() > {while(init()) > {search(); > show(); > } > return 0; > } > Followed by: Post your reply here: |
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